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Is it possible to solve the following integral:

$$\lim\limits_{c \to +\infty} \displaystyle\int_{1}^{1/c} \dfrac{\sin u}{u} \, du$$

using "elementary" methods? By "elementary", I mean those methods that do not involve Complex analysis, Lebesgue Integration, etc (basically, anything beyond an elementary first course in Real Analysis, say, from the first six chapters of Baby Rudin).

I've seen many solutions to this integral (seemingly with different bounds, including the Dirichlet integral), but all of them seem to use methods that would generally not be accessible to someone with just a basic real analysis course.

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    $\begingroup$ Are any of these particularly useful? -- math.stackexchange.com/questions/5248/… $\endgroup$
    – PrincessEev
    Aug 10, 2020 at 4:02
  • $\begingroup$ @EeveeTrainer Unfortunately, no. None of the answers in that link seem to be based just on elementary methods. $\endgroup$
    – Ricky_Nelson
    Aug 10, 2020 at 4:09
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    $\begingroup$ Did you intend for the lower bound to be $0$, or perhaps $1/c$ if you want to avoid the hole at $0$? $\endgroup$
    – Brian Moehring
    Aug 10, 2020 at 4:13
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    $\begingroup$ The integral has been changed many times. When I left it, it would evaluate to $\pi/2$. At this point, it would evaluate to $-\text{Si}(1)$ which has no way (as far as I know) to be written in terms of more elementary functions and/or constants. If the definition of $\text{Si}(x)$ is allowed, then this is very elementary. Otherwise, it's not, but we're so far afield from where we started, I have no clue what your question actually is... $\endgroup$
    – Brian Moehring
    Aug 10, 2020 at 4:47
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    $\begingroup$ As it stands right now, it is trivial. [$\lim_{c\to+\infty}f(1/c)=f(0)$ for a continuous $f$, which is surely our case.] $\endgroup$
    – metamorphy
    Aug 10, 2020 at 13:41

2 Answers 2

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Note

\begin{align} \int^{\infty}_0 \dfrac{\sin u}{u} \, du &= \int^{\infty}_0 {\sin u}\left(\int_0^\infty e^{-ut}dt\right) du \\ &= \int^{\infty}_0 \left(\int_0^\infty \sin u e^{-ut} du\right)dt\\ & = \int^{\infty}_0 \frac1{1+t^2} dt =\dfrac{\pi}{2} \end{align}

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  • $\begingroup$ @BrianMoehring nevermind, I just saw that OP equated it to $\pi/2$, therefore OP most likely made a typo. $\endgroup$
    – Aniruddha Deb
    Aug 10, 2020 at 4:24
  • $\begingroup$ You are using Fubini's theorem which is from multivariable calculus, but I can't think of any example simpler than this. $\endgroup$
    – Otomeram
    Aug 10, 2020 at 4:31
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    $\begingroup$ How do you justify swapping the integrals here? The form of Fubini's I'm aware of would require $\int_0^\infty \frac{|\sin x|}{x}\,dx$ to be finite, but it's not. $\endgroup$
    – Brian Moehring
    Aug 10, 2020 at 4:33
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    $\begingroup$ @BrianMoehring the integral can be taken up to some finite $R$, and then afterwards justify the limit to infinity. $\endgroup$
    – Ninad Munshi
    Aug 10, 2020 at 4:56
  • $\begingroup$ @NinadMunshi Thanks. The additional work looks fairly ugly at the moment, so I would probably spring for dominated convergence or monotone convergence to make it easier, but I guess there might be an "elementary" way to deal with it. $\endgroup$
    – Brian Moehring
    Aug 10, 2020 at 5:07
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$$\lim_{c\rightarrow\infty^+}\int_1^\frac{1}{c}\frac{\sin(u)}{u}du = \lim_{t\rightarrow0^+}\int_1^t\frac{\sin(u)}{u}du = -\int_0^1\frac{\sin(u)}{u}du = -Si(1)$$

This is the sine integral

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